Work and energy 1. A 10.0-kg crate is pulled 15.0 m up along a frictionless incline as shown in the figure below. The crate starts at rest and has a final speed of 6.00 m/s. motor 15 m 5 a. Draw the free-body diagram for the crate. b. Find the work done by the weight. c. Find the work done by the normal force exerted by the incline on the crate. d. Find the work done by the tension on the rope. You can do this either by using the work-kinetic energy theorem or by using Newton s laws and the definition of work. You should be able to do both.
. A box of mass M = 1.5 kg is pushed against a horizontal spring of constant k = 1000 N/m. The spring is compressed a distance x = 0 cm from its original relaxed position. The box is then released and the box slides on the horizontal surface, which can be considered frictionless. Determine: a. The speed of the box when the spring is still compressed by 10 cm. b. The final speed of the box, when it is not in contact with the spring any more.
3. A small bead of mass m = 30 g hangs at one end of a string of length l = 10 cm. The other end is attached to the ceiling as shown in the figure below. The bead is pulled up, so the string makes an angle 0 = 70 with the vertical, and released from there. Air resistance is negligible. l m a. Is mechanical energy conserved? Why? b. Write the potential energy of the bead as a function of. Take the potential energy to be zero at the lowest point of the trajectory.
c. Fill in the table below with the appropriate values of the mechanical energy, the potential energy, the kinetic energy and the speed of the bead at different angles. E U KE v 70 35 0 70 d. What is the work done by the weight of the bead as the bead moves from the highest to the lowest position? Answers: : 1. b) -6 J c) 0 d) 80 J. a) 4.5 m/s b) 5. m/s 3. c) 19 mj, 19 mj, 0, 0 / 19 mj, 5.3 mj, 14 mj, 0.97 m/s / 19 mj, 0, 19 mj, 1.1 m/s / 19 mj, 19 mj, 0, 0 d) 19 mj
Work and energy 1. A 10.0-kg crate is pulled 15.0 m up along a frictionless incline as shown in the figure below. The crate starts at rest and has a final speed of 6.00 m/s. motor T N 15 m mg 5 a. Draw the free-body diagram for the crate. b. Find the work done by the weight. W mgh 10.0 kg 9.8 m/s 15 m sin 5 60 J (negative because weight points down and vertical displacement is up) c. Find the work done by the normal force exerted by the incline on the crate. 0 (the normal force is perpendicular to the displacement) d. Find the work done by the tension on the rope. You can do this both using the work-kinetic energy theorem or using Newton s laws and the definition of work. You should be able to do both. WKE: W W W W K net g N T 1 60 J 0 WT (10.0 kg) 6.0 m/s 0 W 800 J T Newton s laws: v v0 x 53.4 N15 m 800 J 6.0 m/s T mg sin ma with a 1. m/s 15 m T m a g W T sin 10.0 kg 1. 9.8sin 5 m/s 53.4 N Tx
. A box of mass M = 1.5 kg is pushed against a horizontal spring of constant k = 1000 N/m. The spring is compressed a distance x = 0 cm from its original relaxed position. The box is then released and the box slides on the horizontal surface, which can be considered frictionless. Determine: a. The speed of the box when the spring is still compressed by 10 cm. The work done by a spring as it moves from x0 to x1 is x1 1 1 0 x0 1 1000 N/m 0.1 m 0. m 15 J W kxdx k x x W Work-energy theorem: ` W 1 mv 0 W 15 J v 4.5 m/s m 1.5 kg b. The final speed of the box, when it is not in contact with the spring any more. The block will lose its contact with the spring when it moves faster than the spring. This will happen right after the system achieves its maximum velocity (then the spring will slow down but the box will not). Maximum velocity means maximum kinetic energy, and therefore minimum elastic potential energy. This happens when the spring is in the relaxed position, x=0. W 0 1 0 x0 1 1000 N/m 0. m 0 J W kxdx kx W 0 J v 5. m/s m 1.5 kg
3. A small bead of mass m = 30 g hangs at one end of a string of length l = 10 cm. The other end is attached to the ceiling as shown in the figure below. The bead is pulled up, so the string makes an angle 0 = 70 with the vertical, and released from there. Air resistance is negligible. l m a. Is mechanical energy conserved? Why? Yes. The only force doing work is gravity, which is a conservative force. (The tension on the string does zero work because it is always perpendicular to the trajectory). b. Write the potential energy of the bead as a function of. Take the potential energy to be zero at the lowest point of the trajectory. U( ) mgl(1 cos ) (for details, see lecture notes) c. Fill in the table below with the appropriate values of the mechanical energy, the potential energy, the kinetic energy and the speed of the bead at different angles. mgl 3 (30 10 kg)(9.8 m/s )(10 10 m) 0.094 J.94 mj U ( ) (.94 mj)(1 cos ) E U(70 ) 19 mj (because the initial kinetic energy is zero) and always has the same value. K E U ( ) for any other
E U KE v 70 19 mj 19 mj 0 0 35 19 mj 5.3 mj 14 mj 0.97 m/s 0 19 mj 0 19 mj 1.1 m/s 70 19 mj 19 mj 0 0 d. What is the work done by the weight of the bead as the bead moves from the highest to the lowest position? Wg U U 19 mj bottom U top 0 19 mj (or: W mgh, with h l(1 cos 70 ) 6.6 cm)